### a = 2 - Mr. Roche`s Wiki

```Algebra
Chapter 1
LCH GH
A Roche
p.3
Simplify (i)
=
=
multiply each part by x
factorise the top
Divide top & bottom by (x-3)
=
p.3
Simplify (ii)
+
+
Multiply second part above and below by -1
So that both denominators are the same
Factorise the top
p.3
Simplify
1.
4x(3x2 + 5x + 6) – 2(10x2 + 12x)
4x(3x2 + 5x + 6) – 2(10x2 + 12x)
= 12x3 + 20x2 + 24x – 20x2 - 24x
= 12x3
p.3
Simplify
2.
(x + 2)2 + (x - 2)2 - 8
(x + 2)2 + (x - 2)2 - 8
Expand the squares
= (x2 + 4x + 4) + (x2 - 4x + 4) - 8
= 2x2
p.3
Simplify
3.
(a + b)2 - (a - b)2 – 4ab
(a + b)2 - (a - b)2 – 4ab
Expand the squares
= (a2 + 2ab + b2) - (a2 – 2ab + b2) – 4ab
= a2 + 2ab + b2 - a2 + 2ab - b2 – 4ab
=0
p.3
Simplify
4.
(2a + b)2 – 4a(a + b)
(2a + b)2 – 4a(a + b)
Expand
= (4a2 + 4ab + b2) - 4a2 – 4ab
= 4a2 + 4ab + b2 - 4a2 – 4ab
= b2
p.3
Factorise
5.
x2 + 3x
x2 + 3x
= x(x + 3)
HCF
Factorise
6.
3xy – 6y2
3xy – 6y2
= 3y(x - 2y)
HCF
p.3
Factorise
7.
a2b + ab2
a2b + ab2 = ab(a + b)
HCF
Factorise
8.
9x2 – 16y2
9x2 – 16y2 = (3x – 4y)(3x + 4y) Difference of 2
squares
p.3
Factorise
9.
121p2 – q2
121p2
–
q2 =
(11p – q)(11p + q)
Difference of 2
squares
Factorise
10.
1 – 25a2
1 – 25a2
= (1 – 5a)(1 + 5a)
Difference of 2
squares
p.3
Factorise
11.
x2 – 2x - 8
x2 – 2x - 8
= (x +2)(x
)(x - 4))
Which
factors
to -2?
-8
(1)(-8)
(2)(-4) 
(4)(-2)
(8)(-1)
Factorise
12.
3x2 + 13x - 10
3x2 + 13x - 10
+15x
= (3x – 2)(x
)(x + 5))
-2x
Check!
Factors
-10
(1)(-10)
(2)(-5)
(5)(-2)
(10)(-1)
p.3
Factorise
13.
6x2 - 11x + 3
6x2 - 11x + 3
-9x
= (3x
( – 1)(2x
)(2x
)( - 3)))
-2x
Check!
Factors
6x2
(6x)(x)
(3x)(2x) 
+3
(1)(3)
(-1)(-3) 
p.7 Example
(i) If a(x + b)2 + c = 2x2 + 12x + 23,
for all x, find the value of a, of b and of c.
a(x + b)2 + c = 2x2 + 12x + 23
Expand the LHS
a(x2 + 2xb + b2) + c = RHS
Observe that the LHS is a
ax2 + 2axb + ab2 + c = RHS
(a)x2 + (2ab)x + (ab2 + c) = 2x2 + 12x + 23
a =2
2ab = 12
2(2)b = 12
4b = 12
b=3
Equate coefficients of like
terms
ab2 +c = 23
(2)(3)2 +c = 23
18 +c = 23
c =5
p.7 Example
(ii) If (ax + k)(x2 – px +1) = ax3 + bx + c,
for all x, show that c2 = a(a – b).
(ax + k)(x2 – px + 1) = ax3 + bx + c
Expand the LHS
ax(x2 –px + 1) +k(x2 –px + 1) = RHS
ax3 - apx2 + ax + kx2 –kpx + k = RHS
(a)x3 + (-ap + k)x2 + (a - kp)x + k = ax3 + 0x2 + bx + c
Equate coefficients of like terms
a= a
k - ap = 0
a – kp = b
c - ap = 0
a – c(c/a) = b
c = ap
p = c/a
a – c2 /a = b
a – b = c2 /a
a(a – b) = c2
k = c
p.9 Example
Write out each of the following in the form ab, where b is prime:
(i) 32
(ii)
45
(iii)
75
(i) 32 = (16 x 2)= 162= 42
(ii) 45 = (9 x 5) = 35
(iii) 75 =(25 x 3) = 53
Divide by the largest
square number:
1
4
9
16
25
36
49
64
81
100
121
144
169
p.9 Example
Express in the form
(iv)
(iv)
(v)
, a, b  N :
(v)
Divide by the largest
square number:
1
4
9
16
25
36
49
64
81
100
121
144
169
p.9 Example
(vi) Express
(vi)
in the form k2.
p.9 Example
(i)
Express 18 + 50 - 8 in the form ab, where b is prime.
(ii) 20 - 5 + 45 = k5; find the value of k.
(i) 18 + 50 - 8
= (9 x 2) + (25 x 2) - (4 x 2)
= 32 + 52 – 22
= 62
(ii)
20 - 5 + 45
= (4 x 5) - 5 + (9 x 5)
= 25 - 5 + 35
= 45
Divide by the largest
square number:
1
4
9
16
25
36
49
64
81
100
121
144
169
P.10
Examples of Compound Surds
a + b
1 + 5
a- b
3- 24
a- b
13- 7
P.10
Conjugate Surds
WhenSurd
a compound
surd
Compound
Conjugate
1
isConjugate 2
multiplied by its conjugate the
a - b
a
+
b
result is a rational number. - a + b
a - b
a + b
-a - b
We use this ‘trick’ to solve
fractions
with
compound
a
- b
- a + b
a + b
denominators
p.10 Example
Show that
Multiply top and bottom by
conjugate of denominator
Note that the bottom is
difference of 2 squares
1–3
Q.E.D.
p.11
Solving Simultaneous Equations
For complicated simultaneous
equations we use the
substitution-elimination
method
Solve for x and y the simultaneous equations:
x + 1 – y + 3 = 4,
2
3
x +1 – y+ 3 = 4
2
3
(6)(x + 1)– (6)(y + 3) = (6)4
2
3
(3)(x + 1)– (2)(y + 3) = 24
3x + 3 – 2y - 6 = 24
3x– 2y = 27
p.12 Example
x +y – 3= 1
2
2
x +y – 3= 1
2
2
Get rid of
fractions by
Multiplying
2x + 2(y – 3) = 2(1)
2
2
2x + y – 3 = 1
2x + y = 4
Now solve these simultaneous equations in the normal way
x = 5 and y = -6
p.12-13 Example
Solve for x, y and z:
x + 2y + z = 3 1
5x – 3y +2z = 19
3x + 2y – 3z = -5
Eliminate z from
2 equations
5x – 3y +2z = 19
-2x - 4y -2z = -6
3x - 7y
= 13
2
1
x -2
2
3
3x + 2y – 3z = -5
3x + 6y + 3z = 9
6x + 8y
4
Now solve simultaneous
equations 4 & 5 in the usual way
Label the equations
1, 2 & 3
=4
3
1
5
We find that: x = 2
y = -1
Sub these values
into equation 1
x + 2y + z = 3
(2) + 2(-1) + z = 3
z=3
x3
p.13
Note:
If one equation contains only 2 variables then the other 2
equations are used to obtain a second equation with the same
two variables
e.g. Solve
3x + 2y - z = -3
5x – 3y +2z = 3
5x + 3z = 14
1
2
3
Here, from equation 1 and 2, y should be eliminated to obtain an
equation in x and z, which should then be used with equation 3
```